Question

A carnor's engine takes 2095 j of heat per cycle from source at 400 k and rejects 1676 j to the sink. calculate the temperature of the sink and efficiency of the engine

Answer 1

for carnot cycle

let Q1 = heat to the engine= 2095j
Q2 = heat rejection = 1676 j
T1= source temp = 400 k
T2 = sink temp

from carnots relations

work done by system = W= Q1 -Q2

=> W = 2095-1676 = 419 J

now

efficiency = W / Q1 = 419 / 2095 = 0.20

n= efficiency = 20%

also

n = (T1 - T2) / T1

0.20 ×400 = 400 -T2

=> T2 = 400-80 = 320 K